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bulgar [2K]
2 years ago
12

Here is a list of numbers: -20.-20, 2,7.-14, -4,-17, 13,7 State the median

Mathematics
2 answers:
Scrat [10]2 years ago
6 0
-5.11 because it’s obvious
alukav5142 [94]2 years ago
5 0

Answer:

-5.11

Step-by-step explanation:

as the sum of these numbers is -46 so divide -46 by how many numbers are there for eg:

if there are 5 numbers

20,30,40,50,60

so find the sum of these numbers and divide by 5

as their are 5 numbers

so in our case we have to divide -46 by 9 as there are 9 numbers so our answer is -5.11

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What is the following product? 3√10(y^2√4 + √8y)
Elenna [48]

Answer:

6y^2√10 +12y √5

Step-by-step explanation:

3√10(y^2√4 + √8y)

Distribute

3√10(y^2√4) +3√10 (√8y)

3y^2 √40+3y√80

We know that √ab = √a√b

3y^2 √4*√10+3y√16√5

3y^2 *2*√10+3y*4√5

6y^2√10 +12y √5

8 0
2 years ago
Read 2 more answers
There are 26 students in an algebra class. The number of girls in the class is one less than twice the number of boys. How many
Vladimir79 [104]

9 boys because

x+(2x-1)=26

3x=27

x=9.

2x-1=17.

7 0
3 years ago
True or false is 6x10^5 is 20 times as much as 3x10^4
natta225 [31]

answer

TRUE

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6 0
2 years ago
Read carefully and choose the name of the student who made the correct statement ..... HELPPP
Taya2010 [7]

Answer:

Nolan is correct.

Step-by-step explanation:

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7 0
3 years ago
A random sample of 64 students at a university showed an average age of 20 years and a sample standard deviation of 4 years. The
adelina 88 [10]

Answer:

The 90%  confidence level is  19.15<  L  <   20.85

Step-by-step explanation:

From the question we are told that

     The sample size is  n =  64

     The mean age is  \= x  =  20  \ years

      The standard deviation  is   \sigma  =  4 \ years

 

Generally  the degree of freedom for this data set is mathematically represented as

        df  = n -  1

substituting values

        df  = 64 -  1

        df  = 63

Given that the level of confidence is  90%  the significance level is mathematically evaluated as

          \alpha  =  100 - 90

         \alpha  =10 %  

         \alpha   = 0.10

Now   \frac{\alpha }{2}  =  \frac{0.10}{2}  = 0.05

Since we are considering a on tail experiment

The  critical value for half of  this significance level at the calculated  degree of freedom is obtained from the critical value table as

           t_{df, \frac{ \alpha}{2}   } = t_{63,  0.05   } =  1.669

   The margin for error is mathematically represented as

          MOE  =  t_{df ,  \frac{\alpha }{2} } *  \frac{\sigma}{\sqrt{n} }

substituting values  

          MOE  = 1.699  *   \frac{4 }{\sqrt{64} }

         MOE  = 0.85

he 90% confidence interval for the true average age of all students in the university is evaluated as follows

           \= x - MOE  <  L  <  \= x  + E

substituting values  

         20  - 0. 85 <  L  <   20  + 0.85

         19.15<  L  <   20.85

4 0
3 years ago
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